Question: $ E = \left[\begin{array}{rr}-2 & 5 \\ 0 & 5 \\ 3 & 2\end{array}\right]$ $ D = \left[\begin{array}{rr}1 & -1 \\ 2 & -1\end{array}\right]$ What is $ E D$ ?
Answer: Because $ E$ has dimensions $(3\times2)$ and $ D$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ E D = \left[\begin{array}{rr}{-2} & {5} \\ {0} & {5} \\ \color{gray}{3} & \color{gray}{2}\end{array}\right] \left[\begin{array}{rr}{1} & \color{#DF0030}{-1} \\ {2} & \color{#DF0030}{-1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{1}+{5}\cdot{2} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{1}+{5}\cdot{2} & ? \\ {0}\cdot{1}+{5}\cdot{2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{1}+{5}\cdot{2} & {-2}\cdot\color{#DF0030}{-1}+{5}\cdot\color{#DF0030}{-1} \\ {0}\cdot{1}+{5}\cdot{2} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{1}+{5}\cdot{2} & {-2}\cdot\color{#DF0030}{-1}+{5}\cdot\color{#DF0030}{-1} \\ {0}\cdot{1}+{5}\cdot{2} & {0}\cdot\color{#DF0030}{-1}+{5}\cdot\color{#DF0030}{-1} \\ \color{gray}{3}\cdot{1}+\color{gray}{2}\cdot{2} & \color{gray}{3}\cdot\color{#DF0030}{-1}+\color{gray}{2}\cdot\color{#DF0030}{-1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}8 & -3 \\ 10 & -5 \\ 7 & -5\end{array}\right] $